If the product sqrt{a^{frac{1}{a}} cdot (2a)^ | vedclass

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Question

(B) Let the given expression be $P = \sqrt{a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let the given expression be $P = \sqrt{a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty} = \frac{8}{27}$.Squaring both sides,we get $P^2 = a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdots = \left(\frac{8}{27}\right)^2 = \frac{64}{729}$.Expressing as powers of $a$ and $2$: $a^{\left(\frac{1}{a} + \frac{1}{2a} + \frac{1}{4a} + \cdots \right)} \cdot 2^{\left(0 \cdot \frac{1}{a} + 1 \cdot \frac{1}{2a} + 2 \cdot \frac{1}{4a} + 3 \cdot \frac{1}{8a} + \cdots \right)} = \frac{64}{729}$.The exponent of $a$ is $\frac{1}{a} (1 + \frac{1}{2} + \frac{1}{4} + \cdots) = \frac{1}{a} \left(\frac{1}{1 - 1/2}\right) = \frac{2}{a}$.The exponent of $2$ is $\frac{1}{2a} + \frac{2}{4a} + \frac{3}{8a} + \cdots = \frac{1}{a} (\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \cdots)$. This is an Arithmetico-Geometric Progression $(AGP)$ with sum $S = \frac{1/2}{1 - 1/2} + \frac{(1/2)(1/2)}{(1 - 1/2)^2} = 1 + 1 = 2$. So the exponent is $\frac{2}{a}$.Thus,$a^{\frac{2}{a}} \cdot 2^{\frac{2}{a}} = (2a)^{\frac{2}{a}} = \frac{64}{729} = \left(\frac{8}{27}\right)^2 = \left(\left(\frac{2}{3}\right)^3\right)^2 = \left(\frac{2}{3}\right)^6$.Comparing $(2a)^{\frac{2}{a}} = (2/3)^6$,we set $\frac{2}{a} = 6 \Rightarrow a = \frac{1}{3}$ and $2a = 2/3 \Rightarrow a = 1/3$.

If the product $\sqrt{a^{\frac{1}{a}} \cdot (2a)^{\frac{1}{2a}} \cdot (4a)^{\frac{1}{4a}} \cdot (8a)^{\frac{1}{8a}} \cdots \infty} = \frac{8}{27}$,then the value of $a$ is:

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